I'm working on this exercise (not homework) and I would gladly welcome some hints for how to solve it!
Exercise: Let $1 \leq p,q \leq \infty$ be conjugate exponents. Let $a=(a_1,a_2,…)$ be a sequence such that $\sum_1^\infty a_n x_n$ converges for all $x=(x_1,x_2,…) \in l^p$. Prove that $a \in l^q$.
My idea is like this: It's sufficient to show that $a \in l^1$ since $l^1 \subset l^2 \subset… $ I define a family of operators $\{T_n \}_{n=1}^\infty$ by $T_n(x) = \sum_{k=1}^n a_k x_k$. It is clear that each $T_n$ is linear, bounded and that $\sup_{n}|T_n(x)| < \infty$ so by the Uniform Boundedness principle we get $\sup_n \| T_n \| < \infty$.
Is this a good approach? If so, I'd be grateful for some guidance on how to proceed to get to the conclusion:
$$\sum_1^\infty |a_k| < \infty$$
Otherwise, steer me in a better direction 🙂
Thanks in advance
Best Answer
Thanks to the comments in my original post I believe I have a solution. Recall that we know that $\sup_n \|T_n\| < \infty$. First let $1 \leq p \leq \infty$ and fix $n \geq 1$.
Consider the sequence $x=\{x_k\}$ defined by $x_k = |a_k|^{q}/a_k$ when $1 \leq k \leq n$ and $x_k=0$ when $k > n$. Then $x\in l^p$ and
$$\| x \|_p = \left( \sum_{k=1}^n |a_k|^{p(q-1)} \right)^{1/p} = \left( \sum_{k=1}^n |a_k|^{p} \right)^{1/p}$$
Furthermore, using that $T_n$ is bounded, we get
$$T_n(x) = \sum_{k=1}^n |a_k|^{q} \leq \|T_n\|\left( \sum_{k=1}^n |a_k|^{p} \right)^{1/p}. $$ From this we get:
$$ \left( \sum_{k=1}^n |a_k|^{p} \right)^{1/q} \leq \|T_n\| \leq \|T \|$$
And if we let $n \rightarrow \infty$ we get
$$ \| a \|_q \leq \|T_n\| < \infty$$
and hence $a \in l^q$.
If $p=1$ we again fix $n\geq 1$ but instead consider the sequence $x=(0,...,0,1,0...)$ (the 1 on the n:th place). Then $x\in l^p$ and $\|x\|_1 = 1$. Furthermore we get
$$ |T(x)| = |a_n| \leq \|T_n\| \leq \|T\| < \infty.$$
Taking supremum over $n$ we get that $a \in l^\infty$.